On Fri, Nov 29, 2002 at 11:48:53AM -0500, Raul Miller wrote:
> On Fri, Nov 29, 2002 at 12:22:32AM -0500, Andrew Pimlott wrote:
> > What you did propose doesn't seem to do what you want. It says that
> > a defeat by the default option can't be weaker than another defeat.
> > It can still be stronger.
> I wanted to say that a defeat by the default option can't be weaker
> than another defeat.
> > b. A defeat (A,X) is weaker than a defeat (B,Y) if A is not
> > the default option and V(A,X) is less than V(B,Y). Also,
> > (A,X) is weaker than (B,Y) A is not the default option and if
> > V(A,X) is equal to V(B,Y) and V(X,A) is greater than V(Y,B).
You could spell it out:
A defeat (A,X) is weaker than a defeat (D,Y) if A is not the default
option and D is the default option.
A defeat (D,X) is weaker than a defeat (D,Y) if D is the default
option, and V(D,X) is less than V(D,Y).
A defeat (A,X) is weaker than a defeat (B,Y) if neither A nor B
are the default option, and V(A,X) is equal to V(B,Y) and V(X,A)
is greater than V(Y,B).
A defeat (A,X) is weaker than a defeat (B,Y) if neither A nor B are the
default option, and V(A,X) is less than V(B,Y).
A defeat (A,X) is weaker than a defeat (B,Y) if neither A nor B
are the default option, and V(A,X) is equal to V(B,Y) and V(X,A)
is greater than V(Y,B).
The catch is you should be looking at the V(...) stuff only if either
both A and B are the default option, or if neither of them are. I hate
writing these things in programming languages too :(
Maybe you could say:
The strength of defeats are determined as follows:
(1) defeats by non-default options are always weaker than defeats
by the default option; otherwise
(2) a defeat (A,X) is weaker than a defeat (B,Y) is V(A,X) < V(B,Y);
otherwise
(3) a defeat (A,X) is weaker than a defeat (B,Y) when V(X,A) > V(Y,B);
otherwise
(4) the defeats are equal
or something similar, with the idea that each rule is checked in both
directions before moving on...
> A has a 2:1 supermajority requirement, B has no special majority
> requirement, D is the default option, votes are
> 3 ABD
> 1 BDA
> 1 DBA
> A defeats B by 4:1
> B defeats D by 4:1
> D defeats A by 4:3
Actually, A defeats B by 3:2. You meant:
3 ABD
1 BDA
1 DAB
> Which means that A is not in the Schwartz set, but B and D are. So this
> is a tie between B and D, and the person with the casting vote chooses
> whether to settle for B or whether it's worth discussing this more and
> holding another election.
How interesting. So they're not as equivalent as I'd thought. Hrm.
Cheers,
aj
--
Anthony Towns <aj@humbug.org.au> <http://azure.humbug.org.au/~aj/>
I don't speak for anyone save myself. GPG signed mail preferred.
``If you don't do it now, you'll be one year older when you do.''
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